operations with complex numbers...help please?

[tphood]

(5-2i)[sup:1ppx0whp]2[/sup:1ppx0whp] = (25 ? +4i[sup:1ppx0whp]2[/sup:1ppx0whp])

There should be a middle term in place of the question mark in the second expression, however i cannot figure out where it should come from. I can see that 5[sup:1ppx0whp]2[/sup:1ppx0whp] is 25 and -2i[sup:1ppx0whp]2[/sup:1ppx0whp] is 4i[sup:1ppx0whp]2[/sup:1ppx0whp] and i know there is a middle term, but where it comes from i do not know. Please help me understand.

tphood

 

[masters]

(5-2i)[sup:2znlxh6j]2[/sup:2znlxh6j] = (25 ? +4i[sup:2znlxh6j]2[/sup:2znlxh6j])

There should be a middle term in place of the question mark in the second expression, however i cannot figure out where it should come from. I can see that 5[sup:2znlxh6j]2[/sup:2znlxh6j] is 25 and -2i[sup:2znlxh6j]2[/sup:2znlxh6j] is 4i[sup:2znlxh6j]2[/sup:2znlxh6j] and i know there is a middle term, but where it comes from i do not know. Please help me understand.

tphood

Hi tphood,

Look at it this way: \(\displaystyle (5-2i)^2=(5-2i)(5-2i)=5(5-2i)-2i(5-2i)=25-10i-10i+4i^2=25-20i-4=21-20i\)

 

[garf]

Hi Typehood!
Remember that if you square a complex number, you have the square of a binomial. Just be careful with terms which have i multiplied by i.
i*i=-1
i*i*i=-i
i*i*i*i=1
etc.
garf

 

[tphood]

thank you very much...now i see that this should have been easy since i have already been through binomials..

tphood